Main content
Integral Calculus
Course: Integral Calculus > Unit 1
Lesson 9: Fundamental theorem of calculus and definite integralsProof of fundamental theorem of calculus
The fundamental theorem of calculus is very important in calculus (you might even say it's fundamental!). It connects derivatives and integrals in two, equivalent, ways:
The first part says that if you define a function as the definite integral of another function f, then the new function is an antiderivative of f.
The second part says that in order to find the definite integral of f between a and b, find an antiderivative of f, call it F, and calculate F, left parenthesis, b, right parenthesis, minus, F, left parenthesis, a, right parenthesis.
The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.
First, we prove the first part of the theorem.
Next, we offer some intuition into the correctness of the second part.
Finally, we prove the second part of the theorem based on the first part.
Want to join the conversation?
the first proof... thought a little about it and came up w/ an possible other way to solve it:
take it from the Riemann definition (right sum approx) by
b=x (upper bound)
d/dx(int[a;b]f(x)dx = d/dx(f(a+dx)dx+f(a+2dx)dx...n times...+f(b)dx)
as f(a+dx)dx+f(a+2dx)dx... etc are not x-dependent, the derivatives of all go to 0, so
d/dx(int[a;b]f(x)dx=d/dx(f(b)dx)=f(b)
seems reasonable for me, as it all relies on x being both, the upper bound (b) and the variable, as if i just put x=b, considering b as "the max{x}" in the defined interval [a;b]; besides it avoids all the f(t)dt mess that seems confusing.
so, i want to discuss this, even (probably), why is it wrong (didn't saw it anywhere), but seems quite intuitive for me, for dF/dx is just the last d(right area)=f(b)dx/dx + the local linearity stuff.(13 votes)
Looks like you did a good job there, but I still can't understand what you did because it's hard to understand it when it's written in that way. Can you, or anyone else, please write it down on a piece of paper and upload a photo of it on the internet so everybody can comprehend what's going on?
Thank you for your addition.(22 votes)
In the mean value theorem for integrals proof Sal uses the fundamental theorem of calculus and here in the first part he uses the mean value theorem. Isn't that a circular argument because it says that MVT is true from FTC and FTC is true from MVT?(12 votes)
Yes.
But MVT can be proved independently from FTC. In fact, the most rigorous proofs are a bunch of 𝜖-𝛿 style computations, not a corollary from some other strong theorem.(8 votes)
I understood the first part of the theorem this way. Integrating from a to x(between a and b) is like moving x-pivot from a to b. Along the way x -pivot moves, we add the y- value that we get from the function f(t) (Since F(t) is an integration.) So the F(t)'s derivative of the moment we enter the arbitrary x into f(t) must be +f(x), because F(x) is adding f(x) at that moment, and the derivative describes the rate of change at that moment. Is my intuition correct?(7 votes)
If i understood well, you might be thinking that the function itself pivotes like in this simulation: https://bit.ly/2JK8toj (Desmos)
Tell my if it wasn't usefull :D(3 votes)
- I'm confused why we needed to prove that d/dx of F is f, when F was defined during creation as the area under f. When we (or Leibnitz I suppose I should say) specified the "width" as dx, doesn't that in itself say that the derivative of F, (the rate at which the area, F, is changing at x) is given by f(x)? Stated another way, by specifying the width as dx, which is negligible in the area calculation by design, doesn't our area change by f(x)? Perhaps this is not proof because it's recursive?(4 votes)
- That's true, but it isn't a rigorous proof. It's good that you have the intuition for it, though!(4 votes)
What about indefinite integrals? Those don't correspond to any area of a curve, yet they equal the anti-derivative of the function. How does that work?(3 votes)
The indefinite integral is basically to have a general antidriative. Since the antiderivative is the integral. The problem is the derivative of f(x) is the same as the derivative of f(x)+1 or plus or minus any constant. So the general form as I called it was given a C to represent you could take the derivative of this where C is any real number, and you would get the original function.
Let me know if that didn't help.(2 votes)
What is that dt and why does it matter?(5 votes)- this is the formal explanation of dx
https://en.wikipedia.org/wiki/Integral#Meaning_of_the_symbol_dx
to be simple, it's the deferential of x and x + delta x(2 votes)
Would there be any critical errors in this analogy? (I came up with this to understand those theorems more intuitively.)
f = F'
f = productivity (speed in producing value)
F = wealth (accumulated value)
C = Inherited wealth
If productivity is static, wealth increases linearly
if productivity increases, wealth increases exponentially
-> INVEST IN EDUCATION!(3 votes)
The basic idea is correct. Wealth would not necessarily increase exponentially, but it would certainly increase at a much faster rate.(1 vote)
This would be a lot easier if different letters were used for the function. Sometimes it's hard to tell if he means f or F.(3 votes)
Where in the AP calc BC playlist is the video for the MVT of definite integrals? I've gone through all the material before this and I haven't seen it yet, but Sal talks about it in the first video here.(2 votes)- Since the MVT requires the second FTC ( based on the video https://www.khanacademy.org/math/ap-calculus-ab/ab-applications-of-integration-new/ab-8-1/v/mean-value-theorem-integrals ), and the first FTC requires the MVT, shouldn't we prove the second FTC before proving the first one?
Also, how exactly is the second FTC based on the first one? I can't see how the first one would be necessary to prove the second one, and if it were to be necessary we would be caught in a circular reasoning.(2 votes)- In the linked video, Sal is pointing out a connection between the MVT and integration. He is not proving the MVT.
To actually prove the MVT doesn't require either fundamental theorem of calculus, only the extreme value theorem, plus the fact that the derivative of a function is 0 at its extrema (when the derivative exists). That should defuse any fears of circular reasoning.(1 vote)
